Sciencemadness Discussion Board

Formation of red Samarium (II) Chloride

ChemTalk - 15-7-2021 at 14:39

We made a short 60-second video showing the formation of the rarely seen red samarium (II) chloride, along with some close-ups of elemental samarium taken with our new macro lens.

We hope you enjoy, let us know how you like it, and if there are any other rare-earth element reactions or elements you would be interested in seeing.

https://www.youtube.com/watch?v=BjBWzd04Qzo

Scott

PS -we heard that samarium (II) sulfate may be even more stable, but we have not been able to produce it yet.

Bedlasky - 16-7-2021 at 01:18

Nice video! If you have THF and iodine, you can make stable solution of SmI2, which is red. You can also make green solution of YbI2 in the same way. Here are some materials:

https://books.google.cz/books?id=QUfQamGsEgwC&printsec=f...

https://pubs.acs.org/doi/10.1021/ja00528a029

Ce(IV) form red complex [CeCl6]2- in conc. HCl (and also chlorine and Ce3+). Heating cause full decomposition to Ce3+ and Cl2.

I read something about Pr(IV) and Tb(IV) complexes with periodate, tellurate and phosphotungstate, which may be interesting and very strong oxidants.

ChemTalk - 16-7-2021 at 07:54

Bedlasky, thanks for the nice comment, and for the suggestions. We'll try to get some THF. I wonder if it would work in Xylene?

I have several rare earth elements, I have to check to see if I have Ytterbium, I might. we have Cerium (IV). Thanks for the links.
Quote: Originally posted by Bedlasky  
Nice video! If you have THF and iodine, you can make stable solution of SmI2, which is red. You can also make green solution of YbI2 in the same way. Here are some materials:

https://books.google.cz/books?id=QUfQamGsEgwC&printsec=f...

https://pubs.acs.org/doi/10.1021/ja00528a029

Ce(IV) form red complex [CeCl6]2- in conc. HCl (and also chlorine and Ce3+). Heating cause full decomposition to Ce3+ and Cl2.

I read something about Pr(IV) and Tb(IV) complexes with periodate, tellurate and phosphotungstate, which may be interesting and very strong oxidants.

vano - 16-7-2021 at 08:32

It isn't possible for xylene. Look at the photo, in solution samarium iodide is connected with five molecules of THF.

330px-Diiodopenta(THF)samarium(II)-3D-balls.png - 35kB

Bedlasky - 16-7-2021 at 11:40

ChemTalk: Other solvents are possible according that book. There are examples of other solvents: HMPA, acetonitrile, isopropyl alcohol, tert-butyl alcohol, heptan-2-ol and DME (but I honestly don't know what they mean by DME, there are several compounds with this abbreviation). It is important that solvent can coordinate to Sm(II).

[Edited on 16-7-2021 by Bedlasky]

[Edited on 16-7-2021 by Bedlasky]

ChemTalk - 16-7-2021 at 13:01

Bedlasky,

Thanks. It is easy to get THF here, they actually sell it on Amazon.

nezza - 29-7-2021 at 10:26

Here's my video of Samarium dissolving in dilute hydrochloric acid.

Attachment: Samarium-1.mp4 (3.8MB)
This file has been downloaded 185 times

ChemTalk - 29-7-2021 at 20:46

thanks Nezza, this is a great video. I think you posted this some time ago? I feel like I've seen it before

You got much more samarium (II) chloride, yours was much more red. I'm not sure why, maybe your hydrochloric acid was more dilute? Or your samarium piece was larger than mine?